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Homework Help: Integration of a natural log

  1. Aug 24, 2004 #1
    [SOLVED] Integration of a natural log

    I am asked to Integrate by parts

    [tex]\int \ln(2x+1) dx[/tex]

    I plug all of that in and I get,

    [tex]{\int \ln(2x+1) dx\}={\mbox{x}\ln(2x+1)}-{\int \frac{2x}{2x+1} \mbox{dx}}[/tex]

    At this point I tried doing another substitution letting u equal 2x, but this just brought me back to the original question.
    After this anything I try to do doesn't lead me anywhere. I know it can be solved because the answer is in the back of the book.
    First, is what I have done so far correct?
    Second, how do I finish the problem?
    Any help you have would be greatly appreciated.

  2. jcsd
  3. Aug 24, 2004 #2


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    Try approaching it like this:

    Let θ = 2x+1, so that your integral is:

    [tex] \int{\ln\left(2x+1\right) dx} = 2\int{\ln\theta d\theta} [/tex]

    Now you can integrate by parts like you did before, with:

    u = ln(θ) --> du = dθ/θ
    dv = dθ --> v = θ

    You should get:

    [tex] 2\left[ \theta\ln\theta - \theta \right] [/tex]

    as your result. Now all you have to do is plug θ = 2x+1 back in to find the final indefinite integral.
  4. Aug 24, 2004 #3


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    jamesrc's suggestion is good. If you want to go ahead do it the way you were, you can actually do the division [itex]\frac{2x}{2x+1}= 1- \frac{1}{2x+1}[/itex]. Of course, to do the integral of [itex]\frac{1}{2x+1}[/itex], make the substition u= 2x+1 (which gets you right back to jamesrc's suggestion).
    Last edited by a moderator: Aug 25, 2004
  5. Aug 25, 2004 #4
    thanks for your help, I hate it when the answer is so obvious and I can't figure it out.

    Thanks, Erik
  6. Aug 25, 2004 #5
    One final question, when you substituted theda, shouldn't the constant in front of the integral be 1/2 instead of 2???

    (using u instead of theda)

  7. Aug 25, 2004 #6
    Yes, it should be 1/2.
  8. Aug 25, 2004 #7


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    Yes, sorry about that. At least that proves that you know what you're doing.
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